If u=log left(x^3+y^3+z^3-3 x y zright),then | vedclass

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(D) Given,$u=\log \left(x^3+y^3+z^3-3 x y z\right)$.We know that $x^3+y^3+z^3-3 x y z = (x+y+z)(x^2+y^2+z^2-x y-y z-z x)$.So,$u = \log(x+y+z) + \log(x

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$3$

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(D) Given,$u=\log \left(x^3+y^3+z^3-3 x y z\right)$.We know that $x^3+y^3+z^3-3 x y z = (x+y+z)(x^2+y^2+z^2-x y-y z-z x)$.So,$u = \log(x+y+z) + \log(x^2+y^2+z^2-x y-y z-z x)$.Now,calculate the partial derivatives:$u_x = \frac{\partial u}{\partial x} = \frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$$u_y = \frac{\partial u}{\partial y} = \frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$$u_z = \frac{\partial u}{\partial z} = \frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$Adding these,we get:$u_x+u_y+u_z = \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$$u_x+u_y+u_z = \frac{3}{x+y+z}$Therefore,$(x+y+z)(u_x+u_y+u_z) = 3$.

If $u=\log \left(x^3+y^3+z^3-3 x y z\right)$,then $(x+y+z)(u_x+u_y+u_z)$ is equal to

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